why this code dumps core?

[v4vijayakumar]

why the following code dumps core? TIA.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *agrv[])
{
int i = 0;
typedef char ca_t[20];
ca_t tokens[10];
char *token;
char *value = "hhjhjkh, ashdkja, dmnas,lkjlk";

token = strtok(value, " ,\t");
while(token != NULL)
{
strcpy(tokens, token);
++i;
token = strtok(value, " ,\t");
}

return 0;
}

 

[Ian Collins]

why the following code dumps core? TIA.

For the same reason as the code in the thread "Why occur the mistake in
runtime about strtok()?" from a couple of days ago.

 

[Nelu]

F> why the following code dumps core? TIA.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *agrv[])
{
int i = 0;
typedef char ca_t[20];
ca_t tokens[10];
char *token;
char *value = "hhjhjkh, ashdkja, dmnas,lkjlk";

token = strtok(value, " ,\t");
while(token != NULL)
{
strcpy(tokens, token);
++i;
token = strtok(value, " ,\t");
}

return 0;
}

You cannot modify a string literal and strtok modifies its argument.

 

[Keith Thompson]

You cannot modify a string literal and strtok modifies its argument.

<PEDANTIC>
strtok modifies the string to which its argument points.
</PEDANTIC>

 

[Nelu]

<PEDANTIC>
strtok modifies the string to which its argument points.
</PEDANTIC>

Good point.

 

[Martin Ambuhl]

why the following code dumps core? TIA.
[...]
char *value = "hhjhjkh, ashdkja, dmnas,lkjlk";

token = strtok(value, " ,\t");

Because strtok modifies the contents of the string pointed to by its
first argument, and value points to a (non-modifiable) string literal.

 

[Keith Thompson]

why the following code dumps core? TIA.
[...]
char *value = "hhjhjkh, ashdkja, dmnas,lkjlk";
token = strtok(value, " ,\t");

Because strtok modifies the contents of the string pointed to by its
first argument, and value points to a (non-modifiable) string literal.

Yes. It's non-modifiable in the sense that attempting to modify it
invokes undefined behavior, which the compiler isn't required to warn
you about.

You should think of string literals as being constant, even though
they're not really, sort of. If you had declared "value" as:

const char *value = "hhjhjkh, ashdkja, dmnas,lkjlk";

the compiler probably would have given you a warning.

 

[Charles Richmond]

F> why the following code dumps core? TIA.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *agrv[])
{
int i = 0;
typedef char ca_t[20];
ca_t tokens[10];
char *token;
char *value = "hhjhjkh, ashdkja, dmnas,lkjlk";

token = strtok(value, " ,\t");
while(token != NULL)
{
strcpy(tokens, token);
++i;
token = strtok(value, " ,\t");
}

return 0;
}

You cannot modify a string literal and strtok modifies its argument.

And if you fix the above problem, you need to pass NULL as the first
argument to "strtok()" *after* the initial call. Otherwise, you will
end up with an infinite loop and without the result you seek.